SN74AXC1T45: SN74AXC1T45DBVR

Rajat Singh

Intellectual 300 points

Part Number: SN74AXC1T45
Other Parts Discussed in Thread: LSF0101

Hi Team,

I am using SN74AXC1T45DBVR in design.

I am converting VCCA= 1V to VCCB= 1.8V

VOH min is 0.6V of Master device

VIH min of SN74 is 0.7V at 1V(VCCI).

Do you think there will any issue in communication with Master?

Can you suggest 1 bit level translator which have Vih min 0.5V.

Many Thanks

Rajat

8 months ago

0 Michael Ikwuyum 8 months ago

TI__Mastermind 34265 points

Hi Rajat,

Unfortunately, the closest I could find was for 0.65 V.

However, you may also see LSF0101 acting as a passive switch i.e. turns on when input is < VCCA and off when VCCA is reached. And can use pull-up resistors to pull the lines to the intended voltage levels, thanks.

Best Regards,

Michael.

0 Clemens Ladisch 8 months ago

Guru 287840 points

CMOS inputs have a very high impedance, so the actual V_OH will be higher, and this will work. See [FAQ] What is the output voltage (VOH or VOL) when the output current is X or the supply voltage is Y?

0 Rajat Singh 8 months ago in reply to Clemens Ladisch

Intellectual 300 points

Hi Clemens,

I have attached here the drive strength of Driver/Master

In our case VCCO is 1.0V

Can you check at what drive strength VOH will be more then 0.6V?

Many Thanks

Rajat

0 Michael Ikwuyum 8 months ago in reply to Rajat Singh

TI__Mastermind 34265 points

H Rajat,

VOH increases as drive strength reduces i.e. VCC-VOH / IOH = ROH.

For example, VOH 0.4 V gives (1 V - 0.4 V) / 12 mA = 50 ohms.

For 8 mA drive strength, VOH = 1 V - (8 mA * 50 ohms) = 0.6 V.

For 6 mA drive strength, VOH = 0.7 V

For 4 mA drive strength, VOH = 0.8 V

For 2 mA drive strength, VOH = 0.9 V, thanks.

Best Regards,

Michael.

0 Rajat Singh 8 months ago in reply to Michael Ikwuyum

Intellectual 300 points

Hi Michael,

Thanks for the update.It seems above logic will work by changing the drive strength.

Many Thanks

Rajat

+1 Clemens Ladisch 8 months ago in reply to Rajat Singh

Guru 287840 points

The input leakage current of the AXC I/O pin is at most 8 µA. So even with the lowest drive strength of 2 mA, the actual V_OH is 1 V − 0.008 mA × (1 V − 0.9 V) / 2 mA = 0.9996 V. (See the FAQ I linked to.)

Logic

Logic forum

SN74AXC1T45: SN74AXC1T45DBVR